Why list monad combines in that order? - list

Question

I was reading about list monads and encountered:
[1,2] >>= \n -> ['a','b'] >>= \ch -> return (n,ch)
it produces
[(1,'a'),(1,'b'),(2,'a'),(2,'b')]
Here's how I understand it:
Implicit parentheses are:
([1,2] >>= \n -> ['a','b']) >>= (\ch -> return (n,ch))
([1,2] >>= \n -> ['a','b']) should give [('a',1),('b',1),('a',2),('b',2)]
because
instance Monad [] where
return x = [x]
xs >>= f = concat (map f xs) -- this line
fail _ = []
so concat (map f xs) is concat (map (\n -> ['a','b']) [1,2]) which should produce [('a',1),('b',1),('a',2),('b',2)] - quite the opposite of the actual output.
Then I don't understand >>= (\ch -> return (n,ch)) part - I think that n here has no sense. That specific reasoning is flawed, could you please explain how that expression([1,2] >>= \n -> ['a','b'] >>= \ch -> return (n,ch)) is computed step-by-step?

Answer 1

Your implicit parentheses are wrong. The way you have it, the n parameter to the first lambda would not be in scope in the return. It is more like:
([1,2] >>= (\n -> ['a','b'] >>= (\ch -> return (n,ch))))
Which becomes:
concatMap (\n -> concatMap (\ch -> [(n,ch)]) ['a','b']) [1,2]

Answer 2

No, you're using wrong parenthesization. It's nested to the right, so that
[1,2] >>= \n -> ['a','b'] >>= \ch -> return (n,ch)
=
[1,2] >>= (\n -> ['a','b'] >>= (\ch -> return (n,ch) ))
=
do { n <- [1,2]
; do { ch <- ['a','b']
; return (n,ch) }}
=
for n in [1,2]: -- pseudocode
for ch in ['a','b']:
return (n,ch)
=
[ r | n <- [1,2], ch <- ['a','b'], r <- [(n,ch)] ] -- return == (:[])
=
[ (n,ch) | n <- [1,2], ch <- ['a','b'] ]
=
pure (,) <*> [1,2] <*> ['a','b'] -- using list applicative
=
[(1,'a'), (1,'b'), (2,'a'), (2,'b')]
and the innermost list is "spinning" the fastest, like in a car's odometer.
You are absolutely right, with the wrong parenthesization the n binding would make no sense. It is precisely for that reason that it must associate to the right, to make the nested binding possible; for the nested computations are the essence of the monad:
[ foo x | x <- xs ] -- functor : amendable computations
[ bar x y | x <- xs AND y <- ys ] -- applicative : combinable computations
[ baz x y | x <- xs, y <- foo x ] -- monad : reinterpretative computations
(and yes, omitting parentheses in learning material is the root of all evil... not really, but still...)

Answer 3

Implicit parentheses are:
([1,2] >>= \n -> ['a','b']) >>= (\ch -> return (n,ch))
This is wrong. \ has the lowest precedence and thus extends to the end of the expression. So the parentheses are:
[1,2] >>= (\n -> ['a','b'] >>= (\ch -> return (n,ch)))
([1,2] >>= \n -> ['a','b']) should give [('a',1),('b',1),('a',2),('b',2)]
As you can see from the above parenthesization ([1,2] >>= \n -> ['a','b']), isn't actually a subexpression of the given expression. But if it were, its result would be ['a', 'b', 'a', 'b']. n is not actually used anywhere in the expressions ['a', 'b'], so there's no way the numbers could appear in the result.
Then I don't understand >>= (\ch -> return (n,ch)) part - I think that n here has no sense.
Given your parenthesization n would indeed be undefined there. However given the proper parenthesization it should be clear where n comes from: We're still inside the \n -> ... function, so n still refers to the argument of that function, namely the current element of the [1, 2] list.

Answer 4

You're wrong about the semantics of concat (map (\n -> ['a','b']) [1,2]). I've done a step by step evaluation of it below.
concat (map (\n -> ['a','b']) [1,2])
concat [(\n -> ['a','b']) 1, (\n -> ['a','b']) 2]
concat [['a','b'], ['a','b']]
['a','b','a','b']
The last bind is what is used to actually make it a list of tuples.